Question

Consider the decomposition of barium carbonate: BaCO3(s)???BaO(s)+CO2(g)

A) Calculate the equilibrium pressure of CO2 at 298 K.

B) Calculate the equilibrium pressure of CO2 at 1300K .

Answer 1

the given reaction is

BaC03 (s) -----> Ba0 (s) + C02 (g)

equilibrium constant Keq = pC02

we know that

at equilbrium

dGo = -RTlnKeq

now consider the given reaction

BaC03 (s) -----> Ba0 (s) + C02 (g)

dHo = dHo products - dHo reactants

dHo = dHo Ba0 + dHo C02 - dHo BaC03

dHo = -548.1 - 393.509 + 1213

dHo = 271.391 kJ/mol

similarly

dSo = dSo products - dSo reactants

dSo = dSo Ba0 + dSo C02 - dSo BaC03

dSo = 70.4 + 213.6 - 112.1

dSo = 171.9 J/K

now we know that

dGo = dHo - TdSo

so

1) AT 298 K

dGo = ( 271.319 x 1000) - ( 298 x 171.9)

dGo = 220.0928 kJ/mol

but

dGo = -RTlnKeq

so

220.0928 x 1000 = -8.314 x 298 x lnKeq

Keq= 2.629 x 10-39

so

pC02 = Keq = 2.629 x 10-39

equilibrium pressure of C02 at 298K is 2.629 x 10-39 atm

2) AT 1300 K

dGo = ( 271.319 x 1000) - ( 1300 x 171.9)

dGo = 47.849 kJ/mol

but

dGo = -RTlnKeq

so

47.849 x 1000 = -8.314 x 1300 x lnKeq

Keq= 0.01195

so

pC02 = Keq = 0.01195

equilibrium pressure of C02 at 1300 K is 0.01195 atm